SRMJEE Mock Test - 4 (Engineering) - JEE MCQ

The value of g at a height R_e from Earth's surface is .


So,

From Stefan's law, the rate at which energy is radiated by the sun at its surface is

P = σ × 4πr²T⁴
(The sun is a perfectly black body as it emits radiations of all wavelengths and so for it e = 1.)
The intensity of this power at earth's surface (under the assumption R >> r₀) is

= 

Here, m = 10 kg
The resultant force acting on the body is

Let the resultant force F makes an angle θ w.r.t. 8 N8 N force,
From figure, tanθ = 6 / 8 = 3 / 4
The resultant acceleration of the body is
 = 1 m/s²
The resultant acceleration is along the direction of the resultant force.
Hence, the resultant acceleration of the body is 1 m/s²  at an angle of tan⁻¹(3/4)  w.r.t. 8 N force.

By the geometric addition of vectors, the sum of the three vectors shown in the figure is zero, since they form the three sides of a triangle.

That is, all the force vectors add up together and get nullified, and thus, the particle is in equilibrium. 

Since the net force on the particle is zero,  remains unchanged.

The work function of a material is the minimum amount of energy needed to release an electron from the surface of that material, it is given by, 

= 

So, for sodium,

T/4 = 0.14

Frequency T = 0.56

f_c = 1 / T = 1 / 0.56

f = 100/56

or

f = 1.79 Hz

The phenomenon in which two waves of the same frequency, travelling in the same direction superimpose with each other, the alternate maxima and minima are formed in the resultant wave, is named as the interference. In this process, the total mechanical energy of the waves is redistributed in that region of superposition.

The current will be equally divided at junction P. The field at the centre due to wires PQ and SR will be equal in magnitude but opposite in the direction, so its effective field will be zero. Similarly, net field due to wires PS and QR is zero. Therefore, the net field at the centre of the loop is zero.

In other words:
For a square loop, current splits at junctions. Magnetic fields from opposite sides (e.g., PQ and SR) are equal and opposite at the center, canceling out. Net field is zero.

In a capillary tube water rises to a height h such that the hydrostatic pressure (hρg) becomes equal to the excess pressure p = 2σ/R, where R is the radius of curvature of the meniscus. A satellite in a stable orbit around the earth is in a state of weightlessness, i.e. g = 0. Thus the hydrostatic pressure becomes zero. Consequently, water will rise to the top of the tube.

Given,
Tangential acceleration at = 10 m s⁻²
Radius of the circular path r = 5 m
Angle made by net acceleration with radial direction θ = 30°
We know that if the net acceleration makes an angle θ with the radial direction then,  where a_t is the tangential acceleration and a_r is the radial acceleration.

Electrode potential of the cell must be +ve for spontaneous reaction.
Zn²⁺ → Zn; E⁰ = -0.76 V
Cu²⁺ → Cu; E⁰ = -0.34 V
Redox reaction is:

E_cell = E⁰_cathode - E⁰_anode
= -034 - (-0.76)
= +0.42 V
E_cell is positive, so the above reaction is feasible.
Hence, option (2) is the correct answer.

• Mutarotation is the process by which the optical rotation of a sugar (like glucose) changes over time when it is dissolved in water. This happens as the sugar interconverts between its different anomeric forms (α and β forms), reaching an equilibrium state where both forms are present in solution, resulting in a constant optical rotation.
• Epimerisation refers to the conversion between epimers, which are sugars that differ in the configuration of only one stereocenter (except for the anomeric carbon).
• Racemisation is the conversion of a compound into a mixture of equal amounts of enantiomers (optical isomers).
• Anomerisation refers to the interconversion between the α and β anomers of a sugar, but mutarotation is the term used for the entire process of change in optical rotation during this interconversion.

Conclusion: The correct answer is: D: mutarotation.

The electrophile attacks on the centre, which is electron-rich. Since benzene posses pi−electron cloud, hence it is susceptible to electrophilic attack.
If benzene is substituted, then an electron-donating group will increase the electron density on the benzene ring; hence it will  facile the electrophile attack, whereas an electron-withdrawing group, if present on the benzene ring, it will decrease the electron density on the benzene ring, hence the electrophile attack will not be easy.
−OH group is an effective electron donor (due to +M effect) as it increases the electron density by involving its lone pair in the ring via resonance.-
Cl  group shows both −I and +M effect, so it is also an electron donor, but the effect is less than the hydroxy group. Thus, the electron density in phenol is higher than chlorobenzene. Methyl group show only +I effect.
Hence, the phenol will be most readily attacked by electrophiles.

Order: Phenol > Toluene > Benzene > Chlorobenzene.

White precipitate obtained is of BaCl₂ , as the Cl⁻ ion's concentration increases due to the addition of HCl , the ionic product becomes more than solubility product and thus, BaCl₂ is precipitated.

• A: Impurities in BaCl₂ – This is incorrect because the formation of the white precipitate is not due to impurities in BaCl₂ but due to the increased concentration of chloride ions from HCl.
• B: Impurities in HCl – This is not the cause of the precipitate. HCl itself does not cause precipitation unless there's an increase in chloride ion concentration.
• C: Precipitation of BaCl₂ – Correct! The BaCl₂ precipitates due to the increased chloride ion concentration from the addition of HCl, causing the ionic product to exceed the solubility product.
• D: Formation of a complex – This is incorrect. The reaction is not due to complex formation but rather due to exceeding the solubility product of BaCl₂.

Reaction:
Cl₂ + 2I⁻ → I₂ + 2Cl⁻

Initial concentration of I⁻ = 0.20 mol/L.

After 20 minutes, the concentration of I⁻ remains 0.20 mol/L, which means that I⁻ has not reacted and there has been no change in the concentration of I⁻.

Analysis:
Since the concentration of I⁻ does not change, the reaction has not occurred, or the rate of the reaction is zero.

If the concentration of I⁻ remains unchanged, it means that no formation of I₂ has taken place.

Rate of Formation of I₂:
The rate of formation of I₂ depends on the consumption of I⁻. If I⁻ concentration remains constant, the rate of formation of I₂ is zero.

So, the correct answer would be None of these, but the rate is zero.

Fluorine cannot form multiple bonds with the central metal atom but oxygen atom can do the same.

Given, f(x) = x³ + 5x + 1
On differentiating w.r.t. x, we get
f'(x) = 3x² + 5 > 0,   _^R
Since, f(x) is an increasing function, so it is one-one.
Since, f(x) is an odd polynomial, so it will give all values of real number.
Hence, f(x) is one-one and onto R.

Given, 

From (1) & (2)
3a = 6
⇒ a = 2

Given,

Multiplication of R₁ by x, R₂ by y and R₃ by z, reduces the given determinant to

Now take xyz common from the column .

Given,

Differentiating the equation on both the sides, we get,

As per the sign convention, at a point at which f′(x) = 0, the sign should change.

At x = 0, the sign is not changing and since it changes from negative to positive, at x = 3a we get the minima.

Now,

Then, at x=3a, we get,

Hence, the coordinates are (3a, 27/4a)

Given 

Here, f(0⁻) = f(0⁺) = f(0), so it is continuous at x = 0

Here, f′(0⁻) = f′(0⁺)

⇒ f(x)  is differentiable at x = 0

Hence, the function is continuous and differentiable in (−∞,∞).

Let the rate of interest be r% per annum.
Let the amount borrowed be Rs. P.
Interest = (PR × 6)/(100)
P + (6PR/100) = (7/4)P
6PR/100 = (3/4)P
6R = 75
R = 12.5%

Let the length of pipe be 100 units.
After being cut, length of the longer part = 70 units and length of the smaller part = 30 units
The difference in lengths = 40 units
% of longer part more than that of the smaller part = (40/30) × 100 = (400/3)%

Let x - 1/x = 5.

Square both sides: (x - 1/x)² = 5² x² - 2 + 1/x² = 25 x² + 1/x² = 27.

Now, square x² + 1/x² to find x⁴ + 1/x⁴: (x² + 1/x²)² = 27² x⁴ + 2 + 1/x⁴ = 729 x⁴ + 1/x⁴ = 729 - 2 = 727.

Thus, the correct answer is A) 727